JEE Main 2026 — 8 April, Evening Shift — Question 36

$\lim _{x \rightarrow \frac{\pi}{2}} \frac{b\left(1-\cos \left(\frac{\pi}{2}-x\right)\right)}{4\left(\frac{\pi}{2}-x\right)^{2}}=\frac{1}{3} \Rightarrow b=\frac{8}{3}$ $\int_{0}^{2}\left|\mathrm{x}^{2}+2 \mathrm{x}-3\right| \mathrm{dx}=\int_{0}^{2}|(\mathrm{x}+3)(\mathrm{x}-1)| \mathrm{dx}$ put $\mathrm{x}-1=\mathrm{t}$ $\int_{-1}^{1}|(\mathrm{t}+4) \mathrm{t}| \mathrm{dt}=\int_{-1}^{0}\left(-\mathrm{t}^{2}-4 \mathrm{t}\right) \mathrm{dt}+\int_{0}^{1}\left(\mathrm{t}^{2}+4 \mathrm{t}\right) \mathrm{dt}$ $\left(\frac{-t^{3}}{3}-2 t^{2}\right)_{-1}^{0}+\left(\frac{t^{3}}{3}+2 t^{2}\right)_{0}^{1}=4$